\$\begingroup\$I'm pretty sure that most answers from here can be trivially modified to answer this question.\$\endgroup\$
– Don ThousandSep 1 '18 at 16:00

\$\begingroup\$@Arnauld No, the output needs to be two numbers (returned or printed). I guess you could convert the decimal number to string, replace . with whitespace and print it.\$\endgroup\$
– DimChtzSep 1 '18 at 16:06

\$\begingroup\$Great point by @JonathanFrech that will get you to 73 bytes. I think the following form would also work, and is also 73 bytes: def g(n,r=0):u,v=n/10,n%10;return u and g([u-~u,u/v+u%v][v&1],r+1)or(n,r)\$\endgroup\$
– mathmandanSep 2 '18 at 21:16

\$\begingroup\$If you use an inline if statement ( IIf(..) ), integer division (\ ), and change the n>10 to n>9, then you can get this down to an immediate window function worth 97 bytes. n=[A1]:While n>9:e=Right(n,1):n=Left(n,Len(n)-1):n=IIf(e/2=e\2,n*2+1,n\e+n Mod e):c=c+1:Wend:?n;c\$\endgroup\$
– Taylor ScottFeb 15 '19 at 19:28

\$\begingroup\$Hello and welcome to PPCG. As it stands, your answer is not conforming to our default I/O, which essentially calls for either full programs or functions. Take a look at other C++ submissions to get a feel for how to construct these wrappers. Also, please do not be discouraged by your one downvote -- not knowing our rule set on your first post is fine and should not be punished. They may remove it if you fix your answer!\$\endgroup\$
– Jonathan FrechSep 1 '18 at 23:37

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